In $ \triangle A B C, \angle A $ is obtuse, $ P B \perp A C, $ and $ Q C \perp A B $. Prove that $ B C^{2}=\left(A C \times C P +A B \times B Q\right) $.

Given:

In \( \triangle A B C, \angle A \) is obtuse, \( P B \perp A C, \) and \( Q C \perp A B \). 

To do:

We have to prove that \( B C^{2}=\left(A C \times C P +A B \times B Q\right) \).
Solution:
 

In $\triangle BPA$, by Pythagoras theorem,

$AB^2=BP^2+PA^2$ 

$BP^2=AB^2-PA^2$......(i)

In $\triangle BPC$, by Pythagoras theorem,

$BC^2=AB^2-PA^2+(PA+AC)^2$

$BC^2=AB^2-PA^2+PA^2+AC^2+2PA\times AC$

$BC^2=AB^2+AC^2+2PA\times AC$.....(ii)

In $\triangle ACQ$, by Pythagoras theorem,

$AC^2=AQ^2+QC^2$ 

$QC^2=AC^2-AQ^2$......(iii)

In $\triangle BCQ$, by Pythagoras theorem,

$BC^2=BQ^2+QC^2$ 

$BC^2=(BA+AQ)^2+AC^2-AQ^2$

$BC^2=AB^2+AQ^2+2AB\times AQ+AC^2-AQ^2$

$BC^2=AB^2+AC^2+2AB\times AQ$.....(iv)

Adding equations (ii) and (iv), we get,

$BC^2+BC^2=AB^2+AC^2+2PA\times AC+AB^2+AC^2+2AB\times AQ$

$2BC^2=2AB^2+2AB\times AQ+2AC^2+2PA\times AC$

$2BC^2=2AB(AB+AQ)+2AC(AC+PA)$

$2BC^2=2AB\times BQ+2AC\times PC$

Dividing by 2 on both sides, we get,

$BC^2=AB\times BQ+AC\times PC$

Hence proved.

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Updated on: 10-Oct-2022

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